What are the solutions to the equation 1x^2+20x=56 ?

The solutions to the equation 1x^2+20x=56 are x=2(sqrt(39)-5),x=-2(5+sqrt(39))

Find the zeros of 1x^2+20x=56

The zeros of 1x^2+20x=56 are x=2(sqrt(39)-5),x=-2(5+sqrt(39))

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1 x^{2} + 20 x = 56

x = 2 (39 - 5), x = - 2 (5 + 39)

Decimal Notation

x = 2.48999 \dots, x = - 22.48999 \dots

Solution steps

1 \cdot x^{2} + 20 x = 56

Refine

x^{2} + 20 x = 56

Move

56

to the left side

x^{2} + 20 x - 56 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 1 \cdot ( - 56 )}{2 \cdot 1}

2 0^{2} - 4 \cdot 1 \cdot (- 56) = 439

x_{1, 2} = \frac{- 20 \pm 4 39}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 20 + 4 39}{2 \cdot 1}, x_{2} = \frac{- 20 - 4 39}{2 \cdot 1}

x = \frac{- 20 + 4 39}{2 \cdot 1} : 2 (39 - 5)

x = \frac{- 20 - 4 39}{2 \cdot 1} : - 2 (5 + 39)

The solutions to the quadratic equation are:

x = 2 (39 - 5), x = - 2 (5 + 39)

x^{2} - x - b = 0

x, 1^{2} + x^{2} = c \cdot p^{2}

x, 5 x^{2} - 7 x y - 3 y^{2} = 0

x, f (x + 1) = (x + 1)^{2} - 1

x, x^{2} + k (4 x + k - 1) + 2 = 0

What are the solutions to the equation 1x^2+20x=56 ?
The solutions to the equation 1x^2+20x=56 are x=2(sqrt(39)-5),x=-2(5+sqrt(39))
Find the zeros of 1x^2+20x=56
The zeros of 1x^2+20x=56 are x=2(sqrt(39)-5),x=-2(5+sqrt(39))